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-BARSIC- [3]
3 years ago
12

The suspended load of a stream _____.

Physics
2 answers:
marshall27 [118]3 years ago
8 0

Answer:

The correct answer is option A, usually consists of fine sand, silt, and clay particles

Explanation:

The suspended load of a stream consists of small particle particularly clay, silts and fine sands. During the process of sediment transportation, sediment at the bottom of a channel gets uplifted and remain suspended due to their small size. A suspended sediment load do not cause erosion as it is not in the frictional contact of the stream bed.

denis23 [38]3 years ago
4 0
<span>the first one ( usually consists of fine sand, silt, and clay particles )

Hope im right!(: </span>
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A toy car accelerates at a constant rate from rest to a speed of 4 m/s in a time of 0.55 s. What was the magnitude of the accele
Korvikt [17]
  • initial velocity=u=0m/s
  • Final velocity=v=4m/s
  • Time=t=0.5s

\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \sf\longmapsto Acceleration=\dfrac{4-0}{0.5}

\\ \sf\longmapsto Acceleration=\dfrac{4}{0.5}

\\ \sf\longmapsto Acceleration=8m/s^2

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3 years ago
Is it possible for a car to be accelerating to the west while it is driving to the east?
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Answer:

Yes

Explanation:

If the acceleration has an opposite direction to the velocity of the car, this means that it is opposed to is motion. Therefore, it is called deceleration, since the car's velocity will decrease until it stops and then will start it moving towards the west.

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3 years ago
What are the characteristics of the radiation emitted by a blackbody? According to Wien's Law, how many times hotter is an objec
jasenka [17]

Answer:

a) What are the characteristics of the radiation emitted by a blackbody?

The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).

The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).

The spectral density energy is related with the temperature and the wavelength (Planck’s law).

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

Explanation:

A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.

The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.

If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:

E = \sigma T^{4}  (1)

Where \sigma is the Stefan-Boltzmann constant and T is the temperature.

The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):

\lambda max = \frac{2.898x10^{-3} m. K}{T}   (2)

Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:

E_{\lambda} = {{8 \pi h c}\over{{\lambda}^5}{(e^{({hc}/{\lambda \kappa T})}-1)}}}  (3)

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.

Equation (2) can be rewrite in terms of T:

T = \frac{2.898x10^{-3} m. K}{\lambda max}   (4)

Case for the object with the blackbody emission spectrum peak in the blue:

Before replacing all the values in equation (4), \lambda max (450 nm) will be express in meters:

450 nm . \frac{1m}{1x10^{9} nm}  ⇒ 4.5x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{4.5x10^{-7}m}

T = 6440 K

Case for the object with the blackbody emission spectrum peak in the red:

Following the same approach above:

700 nm . \frac{1m}{1x10^{9} nm}  ⇒ 7x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{7x10^{-7}m}

T = 4140 K

Comparison:

\frac{6440 K}{4140 K} = 1.55

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

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What evidence did wegener have to support his theory of plate tectonics?
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Answer:

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Prisms separate <br> light, such as that from the Sun, by wavelength
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Sorry I'm so late, but I just took this test and the answer is white (for people who didn't study well ;) )
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